TITLE: $\lim_{n\to\infty} f(t/n)=0$ for every $t$ implies $\lim_{t\to0^+}f(t)=0$ (???)
QUESTION [20 upvotes]: The question was this: Suppose $f$ is continuous on $(0,\infty)$ and for every $t>0$ $$\lim_{n\to\infty}f(t/n)=0.$$Does it follow that $f(t)\to0$ as $t\to0$ from above?
(Evidently it doesn't go without saying: The question is whether the limit exists - of course it equals $0$ if so.)
I couldn't believe I didn't know the answer. Of course if $f$ is not continuous one could concoct a counterexample, but.
I believe the answer is yes. So the remaining questions are vague: Is this obvious to someone for some reason I don't see? Is it something everybody knows? Edit: Maybe it's not trivial; assuming just $f(t/n)\to0$ for almost every $t$ is not enough.
Here's a proof, I think. Suppose not. Then since $f$ is continuous there exist $\epsilon>0$, $t_k\to0$ and $\delta_k>0$ such that $$|f(t)|\ge\epsilon\quad(|t-t_k|<\delta_k).$$
Let $I_k=(t_k-\delta_k,t_k+\delta_k)$ and $$B_k=\bigcup_{n=1}^\infty nI_k,$$where $nS=\{ns:s\in S\}$.
Note that if $0<a<b$ and $t_k<b-a$ then $$[a,b]\cap B_k\ne\emptyset.$$
This allows you to find a nested sequence of intervals $[a_n,b_n]$ such that every element of $[a_n,b_n]$ lies in $B_k$ for at least $n$ values of $k$. So by compactness there exists $t$ such that $t$ lies in infinitely many $B_k$; hence $f(t/k)$ does not tend to $0$.
Chuckle: An example showing that assuming $f(t/n)\to0$ for almost every $t$ is not enough:
Say $t_k\to0$. Choose $\delta_k>0$ so that the $I_k$ are pairwise disjoint and also $$\sum_km((0,A)\cap B_k)<\infty$$for every $A>0$. Let $f$ be a continuous function on $(0,\infty)$ such that $f=0$ everywhere except on $I_k$, where $f$ has a spike of height $1$.
Almost every $t$ lies in only finitely many $B_k$, and for every such $t$ we have $f(t/n)\to0$.

REPLY [10 votes]: This is a standard application of Baire Category Theorem. Considering $f(\frac 1 t)$ we can state this as follows: f continuous on $(0,\infty)$ and $f(nt) \to 0$ as $n \to \infty$ for all t implies $f(t) \to 0$ as $t \to \infty$.  Write $(0,\infty)$ as union of the sets $\{t:|f(nt)| \leq \epsilon$ for all $n \geq  m\}$ where $\epsilon >0$ is fixed. Since $(0,\infty)$ is of second category there exists m and $a<b$ such that $(a,b)$ is contained in $\{t:|f(nt)| \leq \epsilon$ for all $n \geq  m\}$. Let $x>bm$  and $x(\frac 1a -\frac 1 b)>1$ The interval $(\frac x b ,\frac xa)$ has length exceeding 1 so it contains an integer k. Since $k>\frac x b >m$ and $\frac x k \in (a,b)$ it follows that $|f(k \frac x k)| \leq \epsilon$. Thus $|f(x)| \leq \epsilon$ for $x>bm$  and $x(\frac 1a -\frac 1 b)>1$.